Effective descent for modules along faithfully flat ring maps.

Proposition 35.3.9. Let $R \to A$ be a faithfully flat ring map. Then

1. any descent datum on modules with respect to $R \to A$ is effective,

2. the functor $M \mapsto (A \otimes _ R M, can)$ from $R$-modules to the category of descent data is an equivalence, and

3. the inverse functor is given by $(N, \varphi ) \mapsto H^0(s(N_\bullet ))$.

Proof. We only prove (1) and omit the proofs of (2) and (3). As $R \to A$ is faithfully flat, there exists a faithfully flat base change $R \to R'$ such that $R' \to A' = R' \otimes _ R A$ has a section (namely take $R' = A$ as in the proof of Lemma 35.3.6). Hence, using Lemma 35.3.8 we may assume that $R \to A$ has a section, say $\sigma : A \to R$. Let $(N, \varphi )$ be a descent datum relative to $R \to A$. Set

$M = H^0(s(N_\bullet )) = \{ n \in N \mid 1 \otimes n = \varphi (n \otimes 1)\} \subset N$

By Lemma 35.3.7 it suffices to show that $A \otimes _ R M \to N$ is an isomorphism.

Take an element $n \in N$. Write $\varphi (n \otimes 1) = \sum a_ i \otimes x_ i$ for certain $a_ i \in A$ and $x_ i \in N$. By Lemma 35.3.2 we have $n = \sum a_ i x_ i$ in $N$ (because $\sigma ^0_0 \circ \delta ^1_1 = \text{id}$ in any cosimplicial object). Next, write $\varphi (x_ i \otimes 1) = \sum a_{ij} \otimes y_ j$ for certain $a_{ij} \in A$ and $y_ j \in N$. The cocycle condition means that

$\sum a_ i \otimes a_{ij} \otimes y_ j = \sum a_ i \otimes 1 \otimes x_ i$

in $A \otimes _ R A \otimes _ R N$. We conclude two things from this:

1. applying $\sigma$ to the first $A$ we get $\sum \sigma (a_ i) \varphi (x_ i \otimes 1) = \sum \sigma (a_ i) \otimes x_ i$,

2. applying $\sigma$ to the middle $A$ we get $\sum _ i a_ i \otimes \sum _ j \sigma (a_{ij}) y_ j = \sum a_ i \otimes x_ i$.

Part (1) shows that $\sum \sigma (a_ i) x_ i \in M$. Applying this to $x_ i$ we see that $\sum \sigma (a_{ij})y_ i \in M$ for all $i$. Multiplying out the equation in (2) we conclude that $\sum _ i a_ i (\sum _ j \sigma (a_{ij}) y_ j) = \sum a_ i x_ i = n$. Hence $A \otimes _ R M \to N$ is surjective. Finally, suppose that $m_ i \in M$ and $\sum a_ i m_ i = 0$. Then we see by applying $\varphi$ to $\sum a_ im_ i \otimes 1$ that $\sum a_ i \otimes m_ i = 0$. In other words $A \otimes _ R M \to N$ is injective and we win. $\square$

## Comments (8)

Comment #874 by Bhargav Bhatt on

Suggested slogan: One has effective descent for modules along faithfully flat ring maps.

Comment #2832 by Ko Aoki on

Two typos in the proof: "we may assume that $R \to A$ as a section" should be replaced by "we may assume that $R \to A$ has a section", and "(because $\sigma^0_0 \circ \delta^1_0 = \text{id}$ in any cosimplicial object)" should be replaced by "(because $\sigma^0_0 \circ \delta^1_1 = \text{id}$ in any cosimplicial object)"

Comment #4067 by Stephan on

Could you elaborate on the "Hence by the first conclusion we see that..." part? why is $\Sigma_j \sigma(a_ij)yj$ in M?

Comment #4144 by on

The first conclusion was that $\sum \sigma(a_i)x_i$ is in $M$. Apply this to $x_i$ to get $\sum \sigma(a_{ij}) y_j$ in $M$. OK?

Comment #7526 by WhatJiaranEatsTonight on

Could you please tell me how to deduce that $n$ is in the image of $A\otimes_R M\to N$? I cannot see it from the fact that $\sum\sigma(a_i)x_i\in M$ and $\sum _ i a_ i (\sum _ j \sigma (a_{ij}) y_ j) = \sum a_ i x_ i = n$.

Comment #7529 by on

Because $\sum \sigma(a_{ij})y_j$ is in $M$ by the first conclusion applied to $x_i \otimes 1$. But I think we need to rewrite the proof as it is too succint.

There are also:

• 4 comment(s) on Section 35.3: Descent for modules

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